Question

`\rm \: Find \: (dy)/(dx) if \: y = sin \: x \degree`



`\rm \: Hint :y= sin x° = sin( (\pi \: x)/(180) )`



( Convert degree to ratio )


`\: \: \:`
Thanku:)​

Answer 1

Given that:

`\longmapsto{ \large{ \rm{y = \sin \:x \degree}}}`

We know that,

`\dashrightarrow{ \boxed{ \red{ \rm{1 \degree = \left ( (\pi)/(100) \right)^(c) }}}}`

So, using this, above given can be written as,

`{ \large{ \longrightarrow{ \rm{y = \sin \left( (\pi x)/(180) \right) }}}}`

On differentiating both sides w.r.t. x, we get:

`{ \large { \longrightarrow{ \rm{ (d)/(dx)y = (d)/(dx) \sin \left( (\pi x)/(180) \right) }}}}`

We know that,

`{ \dashrightarrow{ \boxed{ \red{ \rm{ (d)/(dx) \sin x = \cos x }}}}}`

So, using the result, we get:

`{ \large{ \longrightarrow{ \rm{ (d)/(dx) = \cos \left( (\pi \: x)/(180) \: \right) (d)/(dx) \left( (\pi \: x)/(180) \right) }}}}`

We know that,

`{ \dashrightarrow{ \boxed{ \red{ \rm{ (d)/(dx)k \: f(x) = k (d)/(dx) \: f(x)}}}}}`

So, using this, we get:

`{ \large{ \longrightarrow{ \rm{ (dy)/(dx) = \cos \left( (\pi \: x)/(180) \right) \: * \: \left( (\pi)/(180) \right) (d)/(dx) x}}}}`

We know that,

`{ \dashrightarrow{ \boxed{ \red{ \rm{ (d)/(dx) x= 1 }}}}}`

So, using this, we get:

`{ \large{ \longrightarrow{ \rm{ (dy)/(dx) = \left( (\pi)/(180) \right) \: \cos \left( (\pi \: x)/(180) \right ) * 1}}}}`

Hence:

`{ \large{ \leadsto{ \green{ \rm{ (dy)/(dx) = \left( (\pi)/(180) \right) \cos \left( (\pi \: x)/(180) \right) }}}}}`

OR

`{ \large{ \leadsto{ \green{ \rm{ (dy)/(dx) = \left( (\pi)/(180) \right) \cos \: x \degree }}}}}`

`\: \:`

Learn More:

`\boxed{\begin{array}c\bf f(x)&\bf(d)/(dx)f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^(2)(x)\\ \\ \sf cot(x)&\sf-{cosec}^(2)(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf√(x)&\sf(1)/(2√(x))\\ \\ \sf log(x)&\sf(1)/(x)\\ \\ \sf{e}^(x)&\sf{e}^(x)\end{array}}`